本题由2023级25班韩丰羽讲解！
题目展示——先探究、再听讲
设函数\( f(x)=x^{2}(e^{x}-a) \)(1)当\( a=0 \)时,求曲线\( y=f(x) \)在\( (1,f(1)) \)处的切线方程.
(2)若\( f(x) \)是增函数,求\( a \)的值.
(3)当\( 0\lt a\lt 1 \)时,设\( x_{0} \)为\( f(x) \)的极大值点,证明\( 0\lt f(x_{0})\lt \displaystyle\frac{4}{e} \).
解题探究——点击直达
答案解析
解: (1)\( a=0 \)时 \( f(x)=x^{2}e^{x}, f'(x)=(x^{2})’e^{x}+x^{2}(e^{x})’=(x^{2}+2x)e^{x} \)\( f(1)=e \) \( f'(1)=3e \)
\(\therefore y=f(x)\)在\( (1,f(1)) \)处切线为 \( y-f(1)=f'(1)(x-1) \) 即 \( 3ex-y-2e=0 \)
(2)\( f'(x)= [x^{2}(e^{x}-a)]’ \) \(=(x^{2})'(e^{x}-a)+(x^{2})(e^{x}-a)’ \)
\(= 2x(e^{x}-a)+x^{2}e^{x} \) \(=x(2e^{x}+xe^{x}-2a) \)
设\( g(x)=2e^{x}+xe^{x}-2a=(2+x)e^{x}-2a \)
\( g'(x)=2e^{x}+xe^{x}+x(e^{x})’ \) \(=(3+x)e^{x} \)
令\(\begin{cases} g'(x)>0 \\ x\in R \end{cases}\) 解得 \( x>-3 \) \(\therefore g(x) \uparrow (-3,+\infty) \) \(\downarrow (-\infty,-3) \)
\( f(x) = x^2(e^x – a), f'(x) = x(2e^x + xe^x – 2a) = x[(2 + x)e^x – 2a] \geq 0 \)
① \( x \geq 0 \)时，
\( (2 + x)e^x – 2a \geq 0, \)即\( 2a \leq (2 + x)e^x \)
设\( B(x) = (2 + x)e^x, \uparrow [0, +\infty) \)
\( B(x) \geq B(0) = 2, \therefore 2a \leq 2, \ a \leq 1 \)
② \( x < 0 \)时
\( (2 + x)e^x – 2a \leq 0, \)即\( 2a \geq (2 + x)e^x \)
\( B(x) = (2 + x)e^x (x < 0) \)
\( B'(x) = (2 + x)e^x + (2 + x)(e^x)’ = (3 + x)e^x \)
令\( \begin{cases} B'(x) = (3 + x)e^x > 0 \\ x < 0 \end{cases} \)
解得\( x \in (-3, 0] \)
\( f(x) = x^2(e^x – a), f'(x) = x(2e^x + xe^x – 2a) = x[(2 + x)e^x – 2a] \geq 0 \)
① \( x \geq 0 \)时，
\( (2 + x)e^x – 2a \geq 0, \)即\( 2a \leq (2 + x)e^x \)
设\( B(x) = (2 + x)e^x, \uparrow [0, +\infty) \)
\( B(x) \geq B(0) = 2, \therefore 2a \leq 2, \ a \leq 1 \)
② \( x < 0 \)时
\( (2 + x)e^x – 2a \leq 0, \)即\( 2a \geq (2 + x)e^x \)
\( B(x) = (2 + x)e^x (x < 0) \)
\( B'(x) = (2 + x)e^x + (2 + x)(e^x)’ = (3 + x)e^x \)
令\( \begin{cases} B'(x) = (3 + x)e^x > 0 \\ x < 0 \end{cases} \)
解得\( x \in (-3, 0] \)
\( \therefore B(x), \uparrow [-3, 0], \downarrow (-\infty, -3) \)
\( x \to -\infty \ \ B(x) \to 0 \)
\( x = 0 \ \ B(0) = 2 \)
\( \therefore B(x) (x \leq 0) \)图象为
\( \therefore 2a \geq 2, a \geq 1 \)
由①②得 \( a = 1 \)
（3）当\(0 < a < 1\)时，设\(x_0\)为\(f(x)\)的极大值
证明\(0 < f(x_0) < \displaystyle \frac{4}{e^2}\)
解：\(f'(x) = xg(x)\) \(g(x) = 2e^x + xe^x – 2a\)
\(g(0) = 2 – 2a > 0\)
\(g(-2) = -2a < 0\)
\(\because g(x)\)在\((-3, +\infty)\)个
\(\therefore\)存在唯一\(x_0 \in (-2, 0)\)
使\(g(x_0) = (x_0 + 2)e^{x_0} – 2a = 0\)
\(\therefore a = \displaystyle \frac{e^{x_0}(x_0 + 2)}{2}\)
\(f(x_0) = x_0^2 (e^{x_0} – a)\)
\(= x_0^2 \left[e^{x_0} – \displaystyle \frac{e^{x_0}(x_0 + 2)}{2}\right] = -\displaystyle \frac{x_0^3 e^{x_0}}{2}\)
设\(h(x) = \displaystyle \frac{x^3 e^x}{2}\) \(x \in (-2, 0)\) \(h'(x)\)
\(h'(x) = -\displaystyle \frac{(3x^2)e^x + x^3(e^x)}{2} = -\displaystyle \frac{(x^2 + 3x^2)e^x}{2} = -\displaystyle \frac{x^2(x + 3)e^x}{2} < 0\)
\(\therefore h(x)\)在\((-2, 0)\)上↓
\(\therefore h(0) < h(x) < h(-2)\)
\(\therefore h(0) < f(x_0) < h(-2)\)
\(0 < f(x_0) < \displaystyle \frac{4}{e^2}\)
得证